Question

Which of the following is(are) correct for concavity of $f\left(x\right)=\sin x+\cos x$ on $[0,2\pi ]?$

• A. Concave downward on $(0,\frac{3\pi }{4})$
• B. Concave downward on $(\frac{7\pi }{4},2\pi )$
• C. Concave upward on $(\frac{7\pi }{4},2\pi )$
• D. Concave upward on $(\frac{3\pi }{4},\frac{7\pi }{4})$

Answer 1

Answer: (A), (B), (D)

Concave upward on $(\frac{3\pi }{4},\frac{7\pi }{4})$

$f\left(x\right)=\sin x+\cos x$
$f^{\prime }\left(x\right)=\cos x-\sin x$
$f^{\prime \prime }\left(x\right)=-\sin x-\cos x$
Now, $\sin x+\cos x=0$
$\Rightarrow x=$ $\{\frac{3\pi }{4},\frac{7\pi }{4}\}$
For concave downward, $f^{\prime \prime }\left(x\right)<0$
$\Rightarrow -(\sin x+\cos x)<0$
$\therefore x\in (0,\frac{3\pi }{4})\cup (\frac{7\pi }{4},2\pi )$
For concave upward, $f^{\prime \prime }\left(x\right)>0$
$\Rightarrow -(\sin x+\cos x)>0$
$\therefore x\in (\frac{3\pi }{4},\frac{7\pi }{4})$